Greco

Black hole in X-Ray Binary. An X-ray binary consists of 2 stars with masses (the accreting compact object) and (the donor). The orbits are circular with radii and centered on the center of mass. (a) Find the orbital period of the binary following the guid...

11 years ago

The flywheel of a motor is connected to the flywheel of an electric generator by a drive belt. The flywheels are of equal size each of radius . While the flywheels are rotating the tension in the upper and lower portions of the drive belt are and respecti...

11 years ago

power is KE=Iù²/2 = 9000•0.8²/2 = 2880 J. (Whitout the children)

11 years ago

a=((m_1-m_2)*g)/(((0.5*m_p)+(m_1+m_2))) t=sqrt((d*(m_p+2*(m_1+m_2)))/((m_1-m_2)*g))

11 years ago

a=((m_1-m_2)*g)/(((0.5*m_p)+(m_1+m_2))) t=sqrt((d*(m_p+2*(m_1+m_2)))/((m_1-m_2)*g))

11 years ago

T=sqrt((4*pi^2*(r_1+r_2)^3)/((m_1+m_2)*G))

11 years ago

yes the answer is 1.47

11 years ago

T=(m*g)/(1+(m*b^2)/(1/2*m*R^2)) ùf=sqrt((2*l)/(1/2*m*R^2)*(m*g)/(1+(m*b^2)/(1/2*m*R^2))) Ôr=m*g+(2*m*b)/pi*((2*l)/(1/2*m*R^2)*(m*g)/(1+(m*b^2)/(1/2*m*R^2)))

11 years ago

Tr=m*g+(2*m*b)/pi*((2*l)/(1/2*m*R^2)*(m*g)/(1+(m*b^2)/(1/2*m*R^2)))

11 years ago

a zillion thx

11 years ago

your answer is 0.981 Mgl/ùÉc Ic=1/2MR2 l=0,6 R=0.2 M=2 ù =300 g=9.81

11 years ago

T=(m_A+m_B+m_C)/(2*sin(theta))*g x=1/m_B*(((l_1+l_2)*(m_A+m_B+m_C))/2-y*m_A)

11 years ago

T=2903.27

11 years ago

rickross, your solutions for a=41.6 which is comes in contrariety, to a=25.is it surerly right??

11 years ago

a) alpha= 2F/mr ccw b) alpha= 2F/mr cw c) alpha= 8F/3mr ccw

11 years ago

rickross still working on it i ll have it soon..

11 years ago

a) alpha= 2F/mr ccw b) alpha= 2F/mr cw c) alpha= 8F/3mr ccw

11 years ago

a=F.(R+r)/(I/R+m.R)=7 f=F-m.a =-2 BUT you write 2 (without the minus (-) sing)

11 years ago

vA=Sqrt(5*G*m1/4*r)=6904.45 m/s

11 years ago

equation is fine..check your parenthesis write ti down this way Sqrt(5*G*m1/4*r1)

11 years ago

Hil your answer should be 8337.92 m/s

11 years ago

m_1: The mass of the merry-go-round m_2: The mass of the sledgehammer v_1: The velocity of the merry-go-round before the collision v_2: The velocity of the sledgehammer before the collision v_1': The velocity of the merry-go-round after the collision v_2'...

11 years ago

nizo your answer is 1111.96

11 years ago

juanpro a=12.83333 f=2.333

11 years ago

excuse me juanpro a=12.83333 f=3.666

11 years ago

not yet..

11 years ago

no it is wrong!!! he ask you "What is the speed V of the rocket after a burn time of 140 seconds (in m/s)? " that is the half of time and half of % . so .. 1500*ln(1/0.750) = 37.91 m/s a=v/t=37.91/140=0.27

11 years ago

sorr that waw 1500*ln(1/0.9750) = 37.91 m/s

11 years ago

yes ROMA V= u*ln(1/(1-p)) a= V/t

11 years ago

always here Hill...if you can help with ruler problem it will pe magnificent

11 years ago

i know this ,i already done it.thx for the efford!

11 years ago

bw Fy=m*g*cos(theta)-m*w^2*r what's r???

11 years ago

i check it r=L/2

11 years ago

I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19

11 years ago

i check them they are ok thx a lot..

11 years ago

I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19

11 years ago

I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19

11 years ago

I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19

11 years ago

a)25 b)2,55 c)9,27

11 years ago

v=113.14 a=0.90

11 years ago

a) I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19

11 years ago

23220

11 years ago

The angle should be less than 45 degrees how is that possible if 45+a/2 is the correct answer then the angle is more than 45 degrees..

11 years ago

i got the green tick so the right answer is a) 45 + (alpha/2) (45 is the max angle without incline plus the a/2 for the incline b) (v_0^2*(1-sin(alpha))/(g*cos^2(alpha))

11 years ago

and if the temperature was 20∘C?

11 years ago

"d=sqrt(m Vi^2/(mg*alpha+k))" whats that alpha in your solution? shouldn't it be d=sqrt(m Vi^2/(mg*+k))?

11 years ago

a) 45+a/2=51 degrees b)(1−sin)*v0^2/(g*cos^2(á))

11 years ago

i got the green tick so here is the solution equation (1) m1gR=0,5m1v1^2+0,5m2v2^2 m2v2-m1v1=0 so equation (2) v2=m1v1/m2 we substitute to equation (1) m1gR=0,5m1v1^2+0,5m2(m1v1/m2)^2 and we solve for v1 Ps i expect a thanks ( xaxaxaxx)

11 years ago

(v_0^2*(1-sin(alpha))/(g*cos^2(alpha))

11 years ago

Hawk, c) The block will move back and reach a second stop somewhere between x=0 and x=x1. is WRONG

11 years ago

|ôP|= 1.83 |á|= 1.84 T= 1.37

11 years ago

how did you find alpha?

11 years ago

"ì(x)=áx, with á= 1.0 m−1" so alpha is 1.0 m^-1 i got a green tick so in this problem alpha is 1.0 m^-1. In mine was m g k R a 1 10 8 2 0.8 and the answer was x1=1.58 t1=0.59

11 years ago

|ôP|=1.308 |á|= 1.478 T= 1.526

11 years ago

|ôP|= 1.308 |á|= 1.478 T= 1.526

11 years ago

x1= 1.58 t1= 0.392

11 years ago

P(final) =P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha)) green tick

11 years ago

but what is the answer for b and in this problem a)po=0.263

11 years ago

b and c

11 years ago

I = m L^2 + (2/3) m R^2 -(k L^2 + mgL)T = .885 d^2T/dt^2 well calculate the coefficient of T on the left e.g. -(7 + 8) T = .885 d^2T/dt^2 .885 d^2T/dt^2 = - 15 T Torque = -15T where T = 5 * pi/180 alpha = d^2T/dt^2 = - 15 T /.885

11 years ago

.885 w^2 = 15 w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period

11 years ago

|ôP|= 1.745 |á|= 1.577 T= 1.477

11 years ago

some help....anybody????

11 years ago

i need b and c for the problem above (20oC etc) as it is i have only one left i found a ) p=0.263

11 years ago

give your values and i ll tell you the answer

11 years ago

fred to the same system but be careful with the units, give me temperatutr and i ll tell you a)

11 years ago

it is midnight here in Greece and i'm feeling realy tired ....last call for help for b and c as i have only one try left..

11 years ago

and vivipop rescues the day (or better night!)

11 years ago

thx a lot vivipop!!!

11 years ago

fred h=10km=10000m and temp.T= 273+20=293, M=0.029 alpha=0.0065 and M=0.029 and the formula for a) is P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))

11 years ago

P(final) = P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))

11 years ago

Goodnight and thank you all for your help !!!

11 years ago

a) 0.256

11 years ago